Minimum Time to Make Rope Colorful: Greedy Group Processing

You have n balloons arranged in a row, each painted a color given by the string colors, and each with a removal cost given by neededTime[i]. The rope looks bad whenever two consecutive balloons share the same color. You need to remove the minimum-cost set of balloons so that no two adjacent balloons have the same color. Return the total minimum time (cost) to achieve this.

This is LeetCode 1578: Minimum Time to Make Rope Colorful, and it is a great introduction to greedy group processing. The pattern it teaches applies to any problem where you need to reduce consecutive duplicates to a single representative while minimizing some cost.

Why this problem matters

Many real problems involve cleaning up sequences by removing duplicates or redundant entries. Think of deduplicating consecutive log entries, compressing runs of repeated characters, or scheduling tasks where back-to-back identical jobs are wasteful. The core pattern is the same: scan groups of consecutive identical elements, decide which to keep and which to discard, and accumulate the cost.

Once you understand greedy group processing here, you can apply the same template to problems like removing duplicate letters, merging consecutive intervals of the same type, or any scenario where you walk through a sequence and handle "runs" of identical values.

The key insight

In each group of consecutive same-color balloons, you always keep the one with the highest removal cost and remove the rest. This is locally optimal AND globally optimal. Why? Because you must remove all but one balloon from each group (otherwise two consecutive same-color balloons remain). Since you are forced to keep exactly one, keeping the most expensive one minimizes the sum of removed costs.

The trick for computing this efficiently: for each group, sum all the costs, then subtract the maximum. The difference is the cost of removing everything except the most expensive balloon.

cost_for_group = sum_of_group - max_of_group

This avoids having to figure out which balloons to remove individually. You just compute two aggregates per group and move on.

The solution

def min_cost(colors: str, needed_time: list[int]) -> int:
    total = 0
    i = 0

    while i < len(colors):
        group_max = 0
        group_sum = 0

        j = i
        while j < len(colors) and colors[j] == colors[i]:
            group_sum += needed_time[j]
            group_max = max(group_max, needed_time[j])
            j += 1

        total += group_sum - group_max
        i = j

    return total

Let's walk through what each piece does.

The outer while loop advances through the string one group at a time. The variable i marks the start of the current group of consecutive same-color balloons.

The inner while loop scans forward from i as long as the color matches colors[i]. It tracks two values: group_sum (the total cost of all balloons in the group) and group_max (the highest individual cost in the group).

After the inner loop finishes, group_sum - group_max gives the minimum removal cost for that group. You keep the most expensive balloon and remove everything else. The difference is exactly the sum of the removed balloons' costs.

Finally, i = j jumps past the entire group so the outer loop starts on the next color.

Visual walkthrough

Let's trace through colors = "aabaa" and neededTime = [1, 2, 3, 4, 1] step by step. Watch how each group of consecutive same-color balloons gets processed independently.

Notice that groups of size 1 (like the "b" at index 2) contribute zero cost because there is nothing to remove. Only groups with two or more consecutive same-color balloons generate removal costs. After processing all groups, the total cost is 2, and the remaining balloons spell "aba" with no consecutive duplicates.

Complexity analysis

Time is O(n) where n is the length of the colors string. Each balloon is visited exactly once by the inner loop. The outer loop does not re-examine balloons because i jumps to j after each group.

Space is O(1). You only use a constant number of variables (total, i, j, group_sum, group_max) regardless of input size. No extra arrays or data structures are needed.

The building blocks

1. Consecutive group scanning

The pattern of walking through a sequence and processing each "run" of identical values:

i = 0
while i < len(arr):
    j = i
    while j < len(arr) and arr[j] == arr[i]:
        j += 1
    process_group(arr, i, j)
    i = j

This two-pointer group scan is the backbone of run-length encoding, consecutive duplicate removal, and any problem that treats runs of identical values as units. The inner loop finds the end of the current run, you process the group from index i to j - 1, and then you jump i to j to start the next run. Every element is touched exactly once.

2. Sum-minus-max aggregation

The pattern of computing the minimum cost to remove all but the best item from a group:

group_sum = sum(values_in_group)
group_max = max(values_in_group)
removal_cost = group_sum - group_max

This works whenever you must keep exactly one item from a set and want to minimize the total cost of discarding the rest. By keeping the item with the highest cost, you minimize the total of what you remove. You will see this same idea in problems about splitting teams, partitioning resources, or selecting one representative from a cluster.

Edge cases

Before submitting, make sure your solution handles these:

• No consecutive duplicates ("abcde"): every group has size 1, so the total cost is 0. Nothing needs to be removed.
• All same color ("aaaaa"): the entire string is one group. You keep the balloon with the highest cost and remove the other four.
• Two elements, same color ("aa" with times [3, 7]): one group, remove the cheaper one. Cost = 3.
• Two elements, different colors ("ab"): no consecutive duplicates. Cost = 0.
• All costs equal ("aaa" with times [5, 5, 5]): keep any one (they all have the same max). Cost = 10.
• Single balloon ("a"): nothing to compare. Cost = 0.
• Long run followed by single elements ("aaabcd"): only the first group "aaa" has removal cost. The rest contribute 0.

From understanding to recall

You have seen how greedy group processing works: scan consecutive same-color balloons, keep the most expensive one in each group, and add up the removal costs using sum - max. The logic is clean and the code is short. But writing it from scratch in an interview requires having the pattern in muscle memory.

The details that trip people up are small but important. Do you use two nested while loops or a single for loop with a running max? Do you reset group_sum and group_max at the start of each group? Do you advance i to j or to j + 1? These are not conceptual gaps. They are recall gaps, and spaced repetition is the most effective way to close them.

Related posts

• Lemonade Change - Another greedy problem about making locally optimal choices
• Best Time to Buy and Sell Stock - Single-pass greedy processing of an array
• Gas Station - Greedy reasoning about when a valid starting point exists

CodeBricks breaks the Minimum Time to Make Rope Colorful problem into its consecutive group scanning and sum-minus-max aggregation building blocks, then drills them independently with spaced repetition. You type each piece from scratch until the pattern is automatic. When a greedy group processing question shows up in your interview, you do not think about it. You just write it.